$f\,^{\prime}(x)=-10x^4+8x^3$ and $f(1)=9$. $f(0) = $
Finding $f(x)$ We have $f'(x)=-10x^4+8x^3$ and we want to find $f(x)$ : $\begin{aligned}f(x)& = \int f'(x)\,dx \\\\ & = \int (-10x^4+8x^3)\,dx \\\\ & = {-2x^5+2x^4} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(1)=9$. Here's what we get when we plug in $1$ : $\begin{aligned}f(1)&={-2(1)^5+2(1)^4} {+ C}\\\\ &={0} {+ C} \end{aligned}$ We are given that this must equal $9$ : $9 = {0} {+ C}$ Solving the equation gives us ${C=9}$. Finding $f(0)$ Now, we have that $f(x)={-2x^5+2x^4} {+9}$. Let's find $f(0)$ by plugging in $0$ : $\begin{aligned}f(0)&=-2(0)^5+2(0)^4 + 9\\\\ &=9 \end{aligned}$ The answer $f(0) = 9$